Trajectory equations

Objective. Eliminate time to derive a Cartesian path.

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Learn: the key idea

The trajectory equation of a projectile is the Cartesian path y in terms of x, obtained by eliminating time t from the horizontal equation x = ut cos θ and the vertical equation y = ut sin θ − ½gt². Substituting t = x/(u cos θ) into the vertical equation gives y = x tan θ − (gx²)/(2u²cos²θ), a parabola in x. This single equation allows direct calculation of height at any horizontal distance without finding time first. For example, it shows immediately that a projectile launched from ground level returns to y = 0 when x = 0 or at the horizontal range.

Worked example

A projectile is launched at speed u and angle θ from ground level. Derive the trajectory equation giving y in terms of x, u, θ and g.

1 $From x = ut cos θ, time t = x \div (u cos θ)$
2 $Substitute into y = ut sin θ - ½gt²$
3 $y = x tan θ - ½g \times x²/(u²cos²θ)$
4 $y = x tan θ - gx²/(2u²cos²θ)$

Answer: y = x tan θ − gx²/(2u²cos²θ)

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Common mistake. Students substitute t incorrectly, forgetting to square the cos θ term in the gravity part, writing (gx²)/(2u²cos θ) instead of cos²θ. Always square both u and cos θ in the denominator of the trajectory equation's second term.

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